Download Derivation of Dirac's Quantum Condition and Uncertainty Principle and more Assignments Chemistry in PDF only on Docsity! CHM 6470 Homework # 2 Due on Wednesday 9/15 1) Dirac’s Quantum condition. Starting with the Poisson bracket { }, f g g ff g x p x p ∂ ∂ ∂ ∂= − ∂ ∂ ∂ ∂ ( ) ( )where and, ,f f x p g g x p= = , We will try to show { } [ ], , f g i f g= − a. Show that { } { }, ,f g g f= − and [ ] [ ], ,f g g f= − b. Show that { } { } { }, , ,f gh f g h g f h= + (you’ve already shown that [ ] [ ] [ ], , ,f gh f g h g f h= + in HW #1) c. Evaluate { },sf gh and show that { }[ ] { } [ ] , , , , s g f h s g f h = Since s,g,f,h can be any function ⇒ [ ] { }, constant ,xf g f g= where { },f g are physical magnitudes and [ ],f g are operators. The magnitude of the proportionality constant is small, and it shows that the correspondence between functions and operators is significant for small size systems. Let us get more information on the proportionality constant, d. Evaluate [ ],f gφ φ , and using the hermiticity of the f and g operators, show that the expectation value of the commutator must be an imaginary number. Since { },f g is a real number, then { }[ ] , , f g f g must be equal to an imaginary number. The units of { },f g are : [ ] [ ][ ] [ ] f g x p × × where “[ ]” refers to “the units of” and the units of the commutator [ ] [ ] [ ]are equal to , f gf g × e. Evaluate [ ] [ ] [ ] [ ] ? f g x p × = × and derive the units of the proportionality constant. (do they look familiar?) So far, we have shown that { } [ ] where, , f g ik f g k= ∈ The last step in our derivation is to obtain the value of the real constant. Unfortunately, the only way of doing this is experimentally! 2) Show that the following set of operators, which define the “momentum representation”, obeys Dirac’s Quantum condition: ˆ ˆ p p p x x i p → = ∂→ = ∂ 3) Derivation of the uncertainty principle. We start by defying ( ) ( )and A A i B Bφ ψ θ ψ= − = − Let us define where h k kφ θ≡ + ∈ ⇒the bracket 2 0h h h= ≥ ⇒ 2| | 0kφ θ+ ≥ . a. | 0k kφ θ φ θ+ + = for kφ θ= − . Show that in this case ( )2 4θ φ φ θ φ φ θ θ+ = b. For | 0k kφ θ φ θ+ + > the equation cannot have any real roots (real roots would mean 2| | 0h = ) Show that in this case ( )2 4θ φ φ θ φ φ θ θ+ < c. Combining a and b you demonstrated that ( )2 4θ φ φ θ φ φ θ θ+ ≤ d. Evaluate φ φ and θ θ . Show that 2Aφ φ| = ∆ and 2Bθ θ| = ∆ e. Evaluate , and φ θ θ φ
